Please help me with 4th sum

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Please find below the solution to the asked query:

In sumation form, L.H.S. can be written as:L.H.S.=r=023 r+1.25Cr.25Cr+2=r=023 r.25Cr.25Cr+2+r=023 25Cr.25Cr+2=25r=123 24Cr-1.25Cr+2+r=023 25Cr.25Cr+2  As r.nCr=n.n-1Cr-1 ;iNow r=023 25Cr.25Cr+2=25C0.25C2+25C1.25C3+25C2.25C4+....+25C23.25C25 ;iiConsider1+x251+1x25=25C0+25C1x+25C2x2+...25C25x2525C0+25C11x+25C21x2+25C31x3...25C251x25As you can see L.H.S. of ii will be obtained by mutliplying and equating coefficient of 1x2Coefficient  of 1x2 in 1+x251+1x25=Coefficient  of 1x2 in 1+x251+x25x25=Coefficient  of 1x2 in 1x251+x50=Coefficient of x23 in 1+x50 x23x25=1x2=50C23r=023 25Cr.25Cr+2=50C23 ;iiir=123 24Cr-1.25Cr+2=24C0.25C3+24C1.25C4+24C2.25C5+....+24C22.25C25 ;ivConsider:1+x241+1x25= 24C0+24C1x+24C2x2+.....+24C24x24 25C0+25C11x+25C2 1x2+25C31x3+..R.H.S. of iv will be obtained by multiplying and collecting coefficient of 1x3=Cofficient of 1x3 in 1+x241+1x25=Cofficient of 1x3 in 1+x241+x25x25=Cofficient of 1x3 in 1+x49x25=Cofficient of x22 in 1+x49=49C22r=123 24Cr-1.25Cr+2=49C22 ;vPutting values from iii and v in i we get:L.H.S.=25 49C22+50C23On comparing with R.H.S. given in the question we get:k=25, λ=22 and μ=232k-λ-μ=50-22-23=5

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